Odd Thermodynamics Equation

My physics book does this First Law of Thermodynamics example problem during the chapter that shows the steps into getting the solution. They say a drill bores a hole in a block (so doing work) and increases the temperature of the block. It then states the equation they're using: ΔU = Q - Wdone by block. They want to know how much work is done by the drill. So therefore, they alter the equation to ΔU = Q + Wdone by drill. So far this makes perfect sense.

Because no heat is exchanged (because there's no temperature differences between the drill and block), the equation becomes ΔU = 0 + Wdone by drill. But then they confuse me by saying ΔU = Wdone by drill = mCΔT. I am totally confused on why they equate ΔU and W to mCΔT, as mCΔT = Q. Q = 0, so mCΔT would also equal zero I'd think. I realize there's a change in temperature (and internal energy) but I didn't think it was due to heat transfer, only work. So then they claim that Wdone by drill = mCΔT.

I've tried for over an hour to find why they did this, but to no avail. I would appreciate any feedback and the time you take to answer :)

My physics book does this First Law of Thermodynamics example problem during the chapter that shows the steps into getting the solution. They say a drill bores a hole in a block (so doing work) and increases the temperature of the block. It then states the equation they're using: ΔU = Q - Wdone by block. They want to know how much work is done by the drill. So therefore, they alter the equation to ΔU = Q + Wdone by drill. So far this makes perfect sense.

Because no heat is exchanged (because there's no temperature differences between the drill and block), the equation becomes ΔU = 0 + Wdone by drill. But then they confuse me by saying ΔU = Wdone by drill = mCΔT. I am totally confused on why they equate ΔU and W to mCΔT, as mCΔT = Q. Q = 0, so mCΔT would also equal zero I'd think. I realize there's a change in temperature (and internal energy) but I didn't think it was due to heat transfer, only work. So then they claim that Wdone by drill = mCΔT.

I've tried for over an hour to find why they did this, but to no avail. I would appreciate any feedback and the time you take to answer :)

Click to expand...

The assumption is that the all work done by the drill produces heat. The measure of this work, then, is the amount of heat produced, which is mCΔT. This is similar to Joule's experiment of churning water with a paddle wheel driven by a falling weight. Joule measured the change in temperature of the water and found that the work done by the falling weight was equal to the increase in internal energy (mCΔT) of the water.

Staff: Mentor

Toe Jailor,

Your confusion is not your fault. They are using a definition of heat capacity that you are unaccustomed to. When we first learn about heat capacity, we are taught that, if heat is transferred to a body at constant volume, since no work is done, the amount of heat is proportional to the temperature rise, such that Q = mCΔT. This is also equal to the change in internal energy, so that ΔU=Q=mCvΔT.

As we progress in our studies of thermo, we learn a more general definition of heat capacity at constant volume:
[tex]C_v=\left(\frac{\partial U}{\partial T}\right)_v[/tex]
(This is per unit mass or per mole)
This is consistent with the earlier definition when no work is done, but applies to all other situations as well. Your book is using this more general definition. In both definitions, Cv has the same numerical value.

Your confusion is not your fault. They are using a definition of heat capacity that you are unaccustomed to. When we first learn about heat capacity, we are taught that, if heat is transferred to a body at constant volume, since no work is done, the amount of heat is proportional to the temperature rise, such that Q = mCΔT. This is also equal to the change in internal energy, so that ΔU=Q=mCvΔT.

As we progress in our studies of thermo, we learn a more general definition of heat capacity at constant volume:
[tex]C_v=\left(\frac{\partial U}{\partial T}\right)_v[/tex]
(This is per unit mass or per mole)
This is consistent with the earlier definition when no work is done, but applies to all other situations as well. Your book is using this more general definition. In both definitions, Cv has the same numerical value.

Chet

Click to expand...

Thank you. Isn't work done in this case though (the answer gives a value greater than zero). The volume of the block changes because of the hold made by the drill. That's considered work right?

Staff: Mentor

Thank you. Isn't work done in this case though (the answer gives a value greater than zero). The volume of the block changes because of the hold made by the drill. That's considered work right?

Click to expand...

The work done on the part of the block that remains (which constitutes our "system" and heats up) is just frictional work that the drill does at the surface. It assumes that all this energy goes into the block, and none goes into the drill bit (which is implicitly assumed to be insulated).

There is another way of looking at this that may make more sense. If we first take as our system the interface between the drill and the bit, this interface has no mass, so the frictional work done on this interface is equal to the heat that leaves it. This heat all enters the block and causes the temperature of the block to rise. So, if our new system is the portion of the block remaining just beneath the interface, there is heat flow that enters the block, and so its internal energy and temperature rises.

The work done on the part of the block that remains (which constitutes our "system" and heats up) is just frictional work that the drill does at the surface. It assumes that all this energy goes into the block, and none goes into the drill bit (which is implicitly assumed to be insulated).

There is another way of looking at this that may make more sense. If we first take as our system the interface between the drill and the bit, this interface has no mass, so the frictional work done on this interface is equal to the heat that leaves it. This heat all enters the block and causes the temperature of the block to rise. So, if our new system is the portion of the block remaining just beneath the interface, there is heat flow that enters the block, and so its internal energy and temperature rises.

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But since the drill isn't gaining temperature, there's no temperature difference between the drill and block. So it's not really gaining internal energy due to heat transfer. I'm sure you're right, I'm still just confused on why you can apply the equation for heat transfer to a problem dealing with work and internal energy change. What exactly is this process/law/application called? So then maybe I can look up a video explaining it.

I'm sure you're right, I'm still just confused on why you can apply the equation for heat transfer to a problem dealing with work and internal energy change.

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The equation

$$
\Delta U = mc \Delta T
$$

is not for heat transfer only. It is valid for any change of equilibrium states, which ever way and inter-states they proceed.

The simple explanation is that the drill does friction work on the remaining material of the block, so its energy can rise even if it receives neither heat nor volume work from the drill. Whenever the energy of material rises, its temperature will rise and the above equation applies.

Staff: Mentor

But since the drill isn't gaining temperature, there's no temperature difference between the drill and block. So it's not really gaining internal energy due to heat transfer. I'm sure you're right, I'm still just confused on why you can apply the equation for heat transfer to a problem dealing with work and internal energy change. What exactly is this process/law/application called? So then maybe I can look up a video explaining it.

Click to expand...

Toe,

You're asking great questions. This is a bit of a tricky problem involving concepts that you probably haven't dealt with yet. I'm going to try to explain it using a slightly different example. Suppose you have a cylindrical bar of metal, and you wrap an electrical heating blanket around it. Then you wrap insulation around the outside of the electrical heating blanket, so that none of the heat can go into the insulation, and all of the heat has to go into the bar. Now, you turn on the electricity. This is analogous to your problem in the sense that heat is being generated at the surface of the bar, and none of the heat can go into the insulation (which is analogous to the drill bit). Now, you leave the heater on for a while, and then shut it off and let the system come to equilibrium. At the end, the bar will be hotter as a result of the cumulative amount of heat that you supplied to the blanket; its internal energy will be higher.

During the process, you had a transient heat transfer situation occurring. The heat flux at the boundary caused the surface temperature to rise. This then provided the temperature driving force for heat transfer to the bar. Even though the insulation did not receive any of the heat, its surface temperature will also rise. But it's thermal conductivity was very low, so no heat enters. Anyhow, the region near the surface of the bar heats up first, and, as time progresses, the temperature profile advances into the bar. All this behavior can be quantified precisely by solving the transient heat conduction equation in the bar. This will give you the temperature profile within the bar at any time. The key points here are that the temperature of the surface rises, and all the heat generated is conducted into the bar. After the heater is turned off, the temperatures within the bar continue to change as a result of continued heat conduction until the bar equilibrates at a new uniform temperature. After all this, the internal energy of the bar will be higher by the cumulative amount of heat that was added. The temperature of the bar will also be higher. If Q represents the cumulative amount of heat that was added, then ΔU=Q=mCvΔT, where the ΔT refers to the change from the initial temperature to the final equilibrium temperature.